If Δr=2r−1(r+1)!1+1r4r3−2nrabc2n−1(n+1)!−1n3(n+1),n∈Nthen ∑r=1n Δr=

∑r=1n Δr=∑r=1n 2r−1∑r=1n (r+1)!1+1r∑r=1n 4r3−2nrabc2n−1(n+1)!−1n3(n+1)=2n−1(n+1)!−1n3(n+1)abc2n−1(n+1)!−1n3(n+1) R1 and R3 are identical =0

If Δr=2r−1(r+1)!1+1r4r3−2nrabc2n−1(n+1)!−1n3(n+1),n∈Nthen ∑r=1n Δr=

∑r=1n Δr=∑r=1n 2r−1∑r=1n (r+1)!1+1r∑r=1n 4r3−2nrabc2n−1(n+1)!−1n3(n+1)=2n−1(n+1)!−1n3(n+1)abc2n−1(n+1)!−1n3(n+1) R1 and R3 are identical =0

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If $Δ_{r} = |\begin{array}{ccc} 2^{r - 1} & \frac{(r + 1)!}{(1 + \frac{1}{r})} & 4 r^{3} - 2 nr \\ a & b & c \\ 2^{n} - 1 & (n + 1)! - 1 & n^{3} (n + 1) \end{array}|, n \in N$ then $\sum_{r = 1}^{n} Δ_{r} =$

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$a (n!) + b \cdot 2^{n} + c$

(n+3)!

abc

answer is C.

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Detailed Solution

$\sum_{r = 1}^{n} Δ_{r} = |\begin{array}{ccc} \sum_{r = 1}^{n} 2^{r - 1} & \sum_{r = 1}^{n} \frac{(r + 1)!}{(1 + \frac{1}{r})} & \sum_{r = 1}^{n} 4 r^{3} - 2 nr \\ a & b & c \\ 2^{n} - 1 & (n + 1)! - 1 & n^{3} (n + 1) \end{array}|$
$= |\begin{array}{ccc} 2^{n} - 1 & (n + 1)! - 1 & n^{3} (n + 1) \\ a & b & c \\ 2^{n} - 1 & (n + 1)! - 1 & n^{3} (n + 1) \end{array}| R_{1} a n d R_{3} a r e i d e n t i c a l = 0$